\( 2.2 e^{x}-(3+e)=-3 e^{1-x} \) 3 Find the first four terms in the Binomial expansion of \( \frac{1}{\left(\frac{1}{\ln e}+e^{\ln (-3 x)}\right)^{2}} \)

We wish to expand   1   ————————————   [1/ln(e) + e^(ln(–3x)]² step‐by‐step. Step 1. Simplify the expression under the square. Notice that ln(e) = 1 so that   1/ln(e) = 1. Also, e^(ln(–3x)) = –3x (provided –3x is within the domain of the logarithm). Thus, the expression becomes   1/[(1 – 3x)²]  or  (1 – 3x)^(–2). Step 2. Find the Binomial expansion of (1 – 3x)^(–2). Recall the general binomial expansion for a power –k:   (1 + t)^(–k) = Σ [n=0 to ∞] (n+k–1 choose n)·(t)ⁿ for |t| < 1. Here, we set k = 2 and let t = –3x. However, it is actually simpler to use the well‐known expansion:   (1 – u)^(–2) = Σ [n=0 to ∞] (n + 1) uⁿ    (for |u| < 1). In our problem u = 3x (note that the original expression is (1 – 3x); the minus sign is already “built in” to the binomial form). Thus,   (1 – 3x)^(–2) = Σ [n=0 to ∞] (n + 1) (3x)ⁿ. Step 3. Write the first four terms of the expansion. For n = 0:  Term₀ = (0 + 1)·(3x)⁰ = 1 For n = 1:  Term₁ = (1 + 1)·(3x)¹ = 2·3x = 6x For n = 2:  Term₂ = (2 + 1)·(3x)² = 3·9x² = 27x² For n = 3:  Term₃ = (3 + 1)·(3x)³ = 4·27x³ = 108x³ Step 4. Write your final answer. The first four terms in the expansion of   1/[(1/ln e + e^(ln(–3x))]² are   1 + 6x + 27x² + 108x³. Remember, this expansion is valid for |3x| < 1, i.e. |x| < 1/3.